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14334414113 is a prime number
BaseRepresentation
bin11010101100110010…
…11100110100100001
31100222222201200111102
431112121130310201
5213324102222423
610330220124145
71015135336535
oct152631346441
940888650442
1014334414113
116096448538
122940696655
131475998b56
1499da87bc5
1558d68dc28
hex35665cd21

14334414113 has 2 divisors, whose sum is σ = 14334414114. Its totient is φ = 14334414112.

The previous prime is 14334414103. The next prime is 14334414121. The reversal of 14334414113 is 31141443341.

14334414113 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 14052442849 + 281971264 = 118543^2 + 16792^2 .

It is a cyclic number.

It is not a de Polignac number, because 14334414113 - 26 = 14334414049 is a prime.

It is not a weakly prime, because it can be changed into another prime (14334414103) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7167207056 + 7167207057.

It is an arithmetic number, because the mean of its divisors is an integer number (7167207057).

Almost surely, 214334414113 is an apocalyptic number.

It is an amenable number.

14334414113 is a deficient number, since it is larger than the sum of its proper divisors (1).

14334414113 is an equidigital number, since it uses as much as digits as its factorization.

14334414113 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 6912, while the sum is 29.

Adding to 14334414113 its reverse (31141443341), we get a palindrome (45475857454).

The spelling of 14334414113 in words is "fourteen billion, three hundred thirty-four million, four hundred fourteen thousand, one hundred thirteen".